Warm-up Quantifiers and the harmonic series Sets Second warmup Induction Bijections. Writing more proofs. Misha Lavrov

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1 Writing more proofs Misha Lavrov ARML Practice 3/16/2014 and 3/23/2014

2 Warm-up Using the quantifier notation on the reference sheet, and making any further definitions you need to, write the following: You can fool all the people some of the time, and some of the people all the time, but you cannot fool all the people all the time.

3 Warm-up Using the quantifier notation on the reference sheet, and making any further definitions you need to, write the following: You can fool all the people some of the time, and some of the people all the time, but you cannot fool all the people all the time. Let P be the set of all people, T the set of all times, and F (p, t) the statement that person p can be fooled at time t. Then ( p P t T : F (p, t)) ( p P t T : F (p, t)) ( p P t T : F (p, t)).

4 Proving things: a case study Problem Prove that the harmonic series diverges

5 Proving things: a case study Problem Prove that the harmonic series diverges. Proof idea Round down to , to , and so on.

6 Key points to hit 1 It s good to be specific about the rounding: n rounds down to 1 chosen so that 2 k 1 < n 2 k. 2 k This makes it easy to show that there are 2 k 1 terms that round down to 1 2 k, contributing a total of 1 2. One possible punchline: diverges, and the harmonic series is at least as large. Better (fewer infinities): The first 2 k terms of the harmonic series total at least 1 + k 2, which can be arbitrarily large.

7 What is divergence, anyway? Say we have the infinite series a 1 + a 2 + a 3 +. We call S n = n k=1 a k the n-th partial sum.

8 What is divergence, anyway? Say we have the infinite series a 1 + a 2 + a 3 +. We call S n = n k=1 a k the n-th partial sum. When all the a k are positive, the infinite series diverges if and only if the sequence of partial sums tends to infinity. This happens iff: The partial sums become arbitrarily large if we take sufficiently many terms. Which is to say, for all M there is an index n such that S n exceeds M. M n : S n > M.

9 What is divergence, anyway? Say we have the infinite series a 1 + a 2 + a 3 +. We call S n = n k=1 a k the n-th partial sum. When all the a k are positive, the infinite series diverges if and only if the sequence of partial sums tends to infinity. This happens iff: The partial sums become arbitrarily large if we take sufficiently many terms. Which is to say, for all M there is an index n such that S n exceeds M. M n : S n > M. When a series contains negative numbers, things are more complicated: e.g.,

10 Proving a dependence We want to prove that M n : S n > M, where S n = n k=1 1 k. How?

11 Proving a dependence We want to prove that M n : S n > M, where S n = n k=1 1 k. How? Let M be any real number.

12 Proving a dependence We want to prove that M n : S n > M, where S n = n k=1 1 k. How? Let M be any real number. Take n = 2 2M.

13 Proving a dependence We want to prove that M n : S n > M, where S n = n k=1 1 k. How? Let M be any real number. Take n = 2 2M. Then S n = ( ) ( 2 2M 1 1 = ) > = M + 1 > M. }{{ 2} 2M times Therefore the harmonic series diverges. ( 1 2 2M M )

14 Proving a dependence We want to prove that M n : S n > M, where S n = n k=1 1 k. How? Let M be any real number. Take n = 2 2M. Then S n = ( ) ( 2 2M 1 1 = ) > = M + 1 > M. }{{ 2} 2M times Therefore the harmonic series diverges. ( 1 2 2M M [Therefore for any M, there is some n such that S n > M, so S n tends to infinity, and therefore the harmonic series diverges.] )

15 3 A number x is even if x = 2y for some y, and odd if x = 2y + 1 for some y. Prove that all numbers are either even or odd. Exercises 1 There are arbitrarily large numbers of the form which are divisible by 7. Rephrase this statement as For all..., there exists... such that.... Then prove it. (Hint: = ) 2 The infinite series converges to π2 6. This is obviously kind of tricky to prove, so we won t. Prove that (Hint: a similar approach works.) What would you need to show to prove that the sequence of partial sums DOES NOT tend to infinity, using the formal definition?

16 Various kinds of mathematical statements x: Odd numbers exist. To prove this, you give an example of an odd number. x: All numbers are equal to themselves. To prove this, you say Let n be any number, and then prove that n = n.

17 Various kinds of mathematical statements x: Odd numbers exist. To prove this, you give an example of an odd number. x: All numbers are equal to themselves. To prove this, you say Let n be any number, and then prove that n = n. x y: See previous slides. x y: There is a number x such that x + y = y for all y. To prove this, you pick an x, and then do the proof.

18 Various kinds of mathematical statements x: Odd numbers exist. To prove this, you give an example of an odd number. x: All numbers are equal to themselves. To prove this, you say Let n be any number, and then prove that n = n. x y: See previous slides. x y: There is a number x such that x + y = y for all y. To prove this, you pick an x, and then do the proof. x y z For all x, there is a y such that (x + y) + z = z for all z....

19 Proving things about sets This is an exercise in unpacking notation. (You should have a reference sheet for all the notation I will use.) For example: Theorem A B A.

20 Proving things about sets This is an exercise in unpacking notation. (You should have a reference sheet for all the notation I will use.) For example: Theorem A B A. Proof. First of all, A B A means for all x A B, x A.

21 Proving things about sets This is an exercise in unpacking notation. (You should have a reference sheet for all the notation I will use.) For example: Theorem A B A. Proof. First of all, A B A means for all x A B, x A. Suppose x A B. Then x A and x B. Therefore x A.

22 Proving things about sets This is an exercise in unpacking notation. (You should have a reference sheet for all the notation I will use.) For example: Theorem A B A. Proof. First of all, A B A means for all x A B, x A. Suppose x A B. Then x A and x B. Therefore x A. Therefore x A B : x A, which means A B A.

23 Exercises in sets Prove the following: 1 A A B. 2 A. 3 A = A. 4 A (A B) B. 5 (A B) (B A) =. 6 Let A B denote (A B) (B A). Prove that (A B) (B C) (A C) =.

24 Second warmup 1 Prove that if 2 is an integer, then it is odd. 2 Prove that if 2 is rational, then it is an integer.

25 Second warmup 1 Prove that if 2 is an integer, then it is odd. Since 1 < 2 < 4, we have 1 < 2 < 4, so 1 < 2 < 2, and therefore 2 is not an integer. Therefore it is true that if 2 is an integer, it is odd. 2 Prove that if 2 is rational, then it is an integer.

26 Second warmup 1 Prove that if 2 is an integer, then it is odd. Since 1 < 2 < 4, we have 1 < 2 < 4, so 1 < 2 < 2, and therefore 2 is not an integer. Therefore it is true that if 2 is an integer, it is odd. 2 Prove that if 2 is rational, then it is an integer. It suffices to prove that 2 is irrational. Suppose 2 = p q, where p and q are integers. Then p 2 = 2q 2. But the highest power of 2 dividing p 2 is even, while the highest power of 2 dividing 2q 2 is odd. This is a contradiction, so 2 cannot be rational.

27 A simple induction proof Theorem For n 4, n! > 2 n. Proof. Let n = 4; then n! = 24 > 16 = 2 n. If n > 4 and (n 1)! > 2 n 1, then n! = n (n 1)! > n 2 n 1 > 2 2 n 1 = 2 n. By induction, we have n! > 2 n for all n 4.

28 The AM-GM inequality Theorem (AM-GM) For real numbers a 1,..., a n 0, if AM = a 1+ +a n n GM = (a 1 a 2 a n ) 1/n, then AM GM. and Proof outline. We prove three things: 1 That AM GM for n = 2. 2 That the n case implies the 2n case. 3 That the n case implies the n 1 case. These implications give us a path to any value of n from the base case of 2 (though this claim needs proof). For example, to prove n = 17, we go By induction, AM GM for all n.

29 The AM-GM inequality 1 Check that AM GM for n = 2. Start with ( a 1 a 2 ) 2 0. This means a 1 + a 2 2 a 1 a 2 0, or a 1+a 2 2 a 1 a 2. 2 Go from n to 2n. 3 Go from n to n 1.

30 The AM-GM inequality 1 Check that AM GM for n = 2. 2 Go from n to 2n. Split the 2n inequality into two halves: a a 2n 2n = a 1 + +a n n + a n+1+ +a 2n n 2 (a 1 a n ) 1/n + (a n+1 a 2n ) 1/n 2 ((a 1 a n ) 1/n (a n+1 a 2n ) 1/n) 1/2 = (a 1 a 2n ) 1/2n. 3 Go from n to n 1.

31 The AM-GM inequality 1 Check that AM GM for n = 2. 2 Go from n to 2n. 3 Go from n to n 1. Let AM = a 1+ +a n 1 n 1, and set a n = AM. Then: AM = a a n (a 1 a n 1 AM) 1/n n AM n (a 1 a n 1 ) AM AM n 1 (a 1 a n 1 ) AM (a 1 a n 1 ) 1/n.

32 Induction exercises 1 Prove that n = n(n+1) 2 by induction on n. 2 (Recall that the Fibonacci numbers are defined by F 0 = 0, F 1 = 1, and F n = F n 1 + F n 2 for n 2.) Prove that F 3n is even for all n. 3 Prove that for all natural numbers n and for all real x, (1 + x) n 1 + nx. (This also holds for all real n 0 when x 1, a fact known as Bernoulli s inequality.) 4 Prove that for n 6, n! > n 3.

33 Proving things with bijections Theorem ( ) ( ) n n = k n k Proof idea. ( n k) counts subsets of {1, 2,..., n} with k elements. ( n n k) counts subsets with n k elements. We can pair these up, by pairing the subset A, where A = k, with the subset {1, 2,..., n} A. Therefore the number of each type of subset is the same. The general technique is to prove X = Y for two sets X, Y by finding a bijection f : X Y.

34 What is a bijection? A bijection must satisfy two constraints: 1 It hits everything: y Y x X : f (x) = y.

35 What is a bijection? A bijection must satisfy two constraints: 1 It hits everything: y Y x X : f (x) = y. Let B be a subset of {1, 2,..., n} of size n k. Then A = {1, 2,..., n} B is a subset of size k such that f (A) = B.

36 What is a bijection? A bijection must satisfy two constraints: 1 It hits everything: y Y x X : f (x) = y. Let B be a subset of {1, 2,..., n} of size n k. Then A = {1, 2,..., n} B is a subset of size k such that f (A) = B. 2 It hits nothing twice: x 1, x 2 X : f (x 1 ) = f (x 2 ) x 1 = x 2.

37 What is a bijection? A bijection must satisfy two constraints: 1 It hits everything: y Y x X : f (x) = y. Let B be a subset of {1, 2,..., n} of size n k. Then A = {1, 2,..., n} B is a subset of size k such that f (A) = B. 2 It hits nothing twice: x 1, x 2 X : f (x 1 ) = f (x 2 ) x 1 = x 2. Let A 1, A 2 be two subsets of size k. If {1,..., n} A 1 = {1,..., n} A 2, then A 1 = A 2. (Exercise!)

38 What is a bijection? A bijection must satisfy two constraints: 1 It hits everything: y Y x X : f (x) = y. Let B be a subset of {1, 2,..., n} of size n k. Then A = {1, 2,..., n} B is a subset of size k such that f (A) = B. 2 It hits nothing twice: x 1, x 2 X : f (x 1 ) = f (x 2 ) x 1 = x 2. Let A 1, A 2 be two subsets of size k. If {1,..., n} A 1 = {1,..., n} A 2, then A 1 = A 2. (Exercise!) A shortcut is to exhibit an inverse: a function f 1 : Y X such that x X : f 1 (f (x)) = x. This is also easy here.

39 Euler s identity on partitions Theorem (Euler) The number of ways to write n as a sum of odd numbers is equal to the number of ways to write n as a sum of distinct numbers. E.g., 7 = 7 7 = 7 = = = = = = = = (Note: these are also known as partitions of n, and the summands are called parts.)

40 Euler s identity on partitions Proof We construct a bijection f from the first kind of partition to the second kind. Let λ be a partition of n into odd parts. For each odd k, let r k be the number of times k occurs in λ.

41 Euler s identity on partitions Proof We construct a bijection f from the first kind of partition to the second kind. Let λ be a partition of n into odd parts. For each odd k, let r k be the number of times k occurs in λ. Write r k as a sum of distinct powers of 2: r k = 2 a k,1 + 2 a k, a k,l(k).

42 Euler s identity on partitions Proof We construct a bijection f from the first kind of partition to the second kind. Let λ be a partition of n into odd parts. For each odd k, let r k be the number of times k occurs in λ. Write r k as a sum of distinct powers of 2: r k = 2 a k,1 + 2 a k, a k,l(k). Then we obtain f (λ) by making the following replacement, for each k: k } + k + {{ + k } k 2 a k,1 + k 2 a k,2 + + k 2 a k,l(k). r k times

43 Euler s identity on partitions Proof We construct a bijection f from the first kind of partition to the second kind. Let λ be a partition of n into odd parts. For each odd k, let r k be the number of times k occurs in λ. Write r k as a sum of distinct powers of 2: r k = 2 a k,1 + 2 a k, a k,l(k). Then we obtain f (λ) by making the following replacement, for each k: k } + k + {{ + k } k 2 a k,1 + k 2 a k,2 + + k 2 a k,l(k). r k times Exercise: describe the inverse of f.

44 Exercises with bijections 1 Prove that ( ) ( n k = n 1 ) ( k + n 1 k 1) using a bijection. 2 The Catalan numbers count the number of ways to parenthesize a 1 + a a n : e.g., for n = 3, we can write ((a 1 + a 2 ) + a 3 ) or (a 1 + (a 2 + a 3 )); for n = 4, one of the possibilities is ((a 1 + (a 2 + a 3 )) + a 4 ). Prove that the Catalan numbers also count the number of paths from (1, 1) to (n, n) which go up or to the right at each step and also stay within region where x y. For n = 3, we have the paths and